/export/starexec/sandbox2/solver/bin/starexec_run_FirstOrder /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES We consider the system theBenchmark. We are asked to determine termination of the following first-order TRS. 0 : [] --> o false : [] --> o gcd : [o * o] --> o if!6220gcd : [o * o * o] --> o if!6220minus : [o * o * o] --> o le : [o * o] --> o minus : [o * o] --> o s : [o] --> o true : [] --> o le(0, X) => true le(s(X), 0) => false le(s(X), s(Y)) => le(X, Y) minus(0, X) => 0 minus(s(X), Y) => if!6220minus(le(s(X), Y), s(X), Y) if!6220minus(true, s(X), Y) => 0 if!6220minus(false, s(X), Y) => s(minus(X, Y)) gcd(0, X) => X gcd(s(X), 0) => s(X) gcd(s(X), s(Y)) => if!6220gcd(le(Y, X), s(X), s(Y)) if!6220gcd(true, s(X), s(Y)) => gcd(minus(X, Y), s(Y)) if!6220gcd(false, s(X), s(Y)) => gcd(minus(Y, X), s(X)) As the system is orthogonal, it is terminating if it is innermost terminating by [Gra95]. Then, by [FuhGieParSchSwi11], it suffices to prove (innermost) termination of the typed system, with sort annotations chosen to respect the rules, as follows: 0 : [] --> yd false : [] --> tc gcd : [yd * yd] --> yd if!6220gcd : [tc * yd * yd] --> yd if!6220minus : [tc * yd * yd] --> yd le : [yd * yd] --> tc minus : [yd * yd] --> yd s : [yd] --> yd true : [] --> tc We use the dependency pair framework as described in [Kop12, Ch. 6/7], with static dependency pairs (see [KusIsoSakBla09] and the adaptation for AFSMs in [Kop12, Ch. 7.8]). We thus obtain the following dependency pair problem (P_0, R_0, minimal, formative): Dependency Pairs P_0: 0] le#(s(X), s(Y)) =#> le#(X, Y) 1] minus#(s(X), Y) =#> if!6220minus#(le(s(X), Y), s(X), Y) 2] minus#(s(X), Y) =#> le#(s(X), Y) 3] if!6220minus#(false, s(X), Y) =#> minus#(X, Y) 4] gcd#(s(X), s(Y)) =#> if!6220gcd#(le(Y, X), s(X), s(Y)) 5] gcd#(s(X), s(Y)) =#> le#(Y, X) 6] if!6220gcd#(true, s(X), s(Y)) =#> gcd#(minus(X, Y), s(Y)) 7] if!6220gcd#(true, s(X), s(Y)) =#> minus#(X, Y) 8] if!6220gcd#(false, s(X), s(Y)) =#> gcd#(minus(Y, X), s(X)) 9] if!6220gcd#(false, s(X), s(Y)) =#> minus#(Y, X) Rules R_0: le(0, X) => true le(s(X), 0) => false le(s(X), s(Y)) => le(X, Y) minus(0, X) => 0 minus(s(X), Y) => if!6220minus(le(s(X), Y), s(X), Y) if!6220minus(true, s(X), Y) => 0 if!6220minus(false, s(X), Y) => s(minus(X, Y)) gcd(0, X) => X gcd(s(X), 0) => s(X) gcd(s(X), s(Y)) => if!6220gcd(le(Y, X), s(X), s(Y)) if!6220gcd(true, s(X), s(Y)) => gcd(minus(X, Y), s(Y)) if!6220gcd(false, s(X), s(Y)) => gcd(minus(Y, X), s(X)) Thus, the original system is terminating if (P_0, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_0, R_0, minimal, formative). We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows: * 0 : 0 * 1 : 3 * 2 : 0 * 3 : 1, 2 * 4 : 6, 7, 8, 9 * 5 : 0 * 6 : 4, 5 * 7 : 1, 2 * 8 : 4, 5 * 9 : 1, 2 This graph has the following strongly connected components: P_1: le#(s(X), s(Y)) =#> le#(X, Y) P_2: minus#(s(X), Y) =#> if!6220minus#(le(s(X), Y), s(X), Y) if!6220minus#(false, s(X), Y) =#> minus#(X, Y) P_3: gcd#(s(X), s(Y)) =#> if!6220gcd#(le(Y, X), s(X), s(Y)) if!6220gcd#(true, s(X), s(Y)) =#> gcd#(minus(X, Y), s(Y)) if!6220gcd#(false, s(X), s(Y)) =#> gcd#(minus(Y, X), s(X)) By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_0, R_0, m, f) by (P_1, R_0, m, f), (P_2, R_0, m, f) and (P_3, R_0, m, f). Thus, the original system is terminating if each of (P_1, R_0, minimal, formative), (P_2, R_0, minimal, formative) and (P_3, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_3, R_0, minimal, formative). We will use the reduction pair processor with usable rules [Kop12, Thm. 7.44]. The usable rules of (P_3, R_0) are: le(0, X) => true le(s(X), 0) => false le(s(X), s(Y)) => le(X, Y) minus(0, X) => 0 minus(s(X), Y) => if!6220minus(le(s(X), Y), s(X), Y) if!6220minus(true, s(X), Y) => 0 if!6220minus(false, s(X), Y) => s(minus(X, Y)) It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: gcd#(s(X), s(Y)) >? if!6220gcd#(le(Y, X), s(X), s(Y)) if!6220gcd#(true, s(X), s(Y)) >? gcd#(minus(X, Y), s(Y)) if!6220gcd#(false, s(X), s(Y)) >? gcd#(minus(Y, X), s(X)) le(0, X) >= true le(s(X), 0) >= false le(s(X), s(Y)) >= le(X, Y) minus(0, X) >= 0 minus(s(X), Y) >= if!6220minus(le(s(X), Y), s(X), Y) if!6220minus(true, s(X), Y) >= 0 if!6220minus(false, s(X), Y) >= s(minus(X, Y)) We orient these requirements with a polynomial interpretation in the natural numbers. We consider usable_rules with respect to the following argument filtering: if!6220gcd#(x_1,x_2,x_3) = if!6220gcd#(x_2x_3) if!6220minus(x_1,x_2,x_3) = if!6220minus(x_2x_3) This leaves the following ordering requirements: gcd#(s(X), s(Y)) >= if!6220gcd#(le(Y, X), s(X), s(Y)) if!6220gcd#(true, s(X), s(Y)) > gcd#(minus(X, Y), s(Y)) if!6220gcd#(false, s(X), s(Y)) >= gcd#(minus(Y, X), s(X)) minus(0, X) >= 0 minus(s(X), Y) >= if!6220minus(le(s(X), Y), s(X), Y) if!6220minus(true, s(X), Y) >= 0 if!6220minus(false, s(X), Y) >= s(minus(X, Y)) The following interpretation satisfies the requirements: 0 = 0 false = 0 gcd# = \y0y1.2 + y0 + y1 if!6220gcd# = \y0y1y2.2 + y1 + y2 if!6220minus = \y0y1y2.y1 le = \y0y1.0 minus = \y0y1.y0 s = \y0.1 + y0 true = 0 Using this interpretation, the requirements translate to: [[gcd#(s(_x0), s(_x1))]] = 4 + x0 + x1 >= 4 + x0 + x1 = [[if!6220gcd#(le(_x1, _x0), s(_x0), s(_x1))]] [[if!6220gcd#(true, s(_x0), s(_x1))]] = 4 + x0 + x1 > 3 + x0 + x1 = [[gcd#(minus(_x0, _x1), s(_x1))]] [[if!6220gcd#(false, s(_x0), s(_x1))]] = 4 + x0 + x1 > 3 + x0 + x1 = [[gcd#(minus(_x1, _x0), s(_x0))]] [[minus(0, _x0)]] = 0 >= 0 = [[0]] [[minus(s(_x0), _x1)]] = 1 + x0 >= 1 + x0 = [[if!6220minus(le(s(_x0), _x1), s(_x0), _x1)]] [[if!6220minus(true, s(_x0), _x1)]] = 1 + x0 >= 0 = [[0]] [[if!6220minus(false, s(_x0), _x1)]] = 1 + x0 >= 1 + x0 = [[s(minus(_x0, _x1))]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_3, R_0, minimal, formative) by (P_4, R_0, minimal, formative), where P_4 consists of: gcd#(s(X), s(Y)) =#> if!6220gcd#(le(Y, X), s(X), s(Y)) Thus, the original system is terminating if each of (P_1, R_0, minimal, formative), (P_2, R_0, minimal, formative) and (P_4, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_4, R_0, minimal, formative). We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows: * 0 : This graph has no strongly connected components. By [Kop12, Thm. 7.31], this implies finiteness of the dependency pair problem. Thus, the original system is terminating if each of (P_1, R_0, minimal, formative) and (P_2, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_2, R_0, minimal, formative). We apply the subterm criterion with the following projection function: nu(if!6220minus#) = 2 nu(minus#) = 1 Thus, we can orient the dependency pairs as follows: nu(minus#(s(X), Y)) = s(X) = s(X) = nu(if!6220minus#(le(s(X), Y), s(X), Y)) nu(if!6220minus#(false, s(X), Y)) = s(X) |> X = nu(minus#(X, Y)) By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_2, R_0, minimal, f) by (P_5, R_0, minimal, f), where P_5 contains: minus#(s(X), Y) =#> if!6220minus#(le(s(X), Y), s(X), Y) Thus, the original system is terminating if each of (P_1, R_0, minimal, formative) and (P_5, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_5, R_0, minimal, formative). We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows: * 0 : This graph has no strongly connected components. By [Kop12, Thm. 7.31], this implies finiteness of the dependency pair problem. Thus, the original system is terminating if (P_1, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_1, R_0, minimal, formative). We apply the subterm criterion with the following projection function: nu(le#) = 1 Thus, we can orient the dependency pairs as follows: nu(le#(s(X), s(Y))) = s(X) |> X = nu(le#(X, Y)) By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_1, R_0, minimal, f) by ({}, R_0, minimal, f). By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed. As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination. +++ Citations +++ [FuhGieParSchSwi11] C. Fuhs, J. Giesl, M. Parting, P. Schneider-Kamp, and S. Swiderski. Proving Termination by Dependency Pairs and Inductive Theorem Proving. In volume 47(2) of Journal of Automated Reasoning. 133--160, 2011. [Gra95] B. Gramlich. Abstract Relations Between Restricted Termination and Confluence Properties of Rewrite Systems. In volume 24(1-2) of Fundamentae Informaticae. 3--23, 1995. [Kop12] C. Kop. Higher Order Termination. PhD Thesis, 2012. [KusIsoSakBla09] K. Kusakari, Y. Isogai, M. Sakai, and F. Blanqui. Static Dependency Pair Method Based On Strong Computability for Higher-Order Rewrite Systems. In volume 92(10) of IEICE Transactions on Information and Systems. 2007--2015, 2009.